Question

The solution of the differential equation $\frac{dy}{dx}=1-\cos (y-x)\cot (y-x)$ is

• A. $x\tan (y-x)=c$
• B. $x=\tan (y-x)+c$
• C. $x=sec(y-x)+c$
• D. $x+\sec (y-x)=c$

Answer 1

Answer: (D)

We have,
$\frac{dy}{dx}=1-\cos (y-x)\cot (y-x)$
Put $y-x=v$
$\Rightarrow \frac{dy}{dx}=1+\frac{dv}{dx}$
$\therefore 1+\frac{dv}{dx}=1-\cos v\cot v$
$\Rightarrow \frac{dv}{dx}=-\frac{{\cos }^{2}v}{\sin v}$
$\Rightarrow -\int \frac{\sin v}{{\cos }^{2}v}dv=\int dx$
$\Rightarrow -\int \sec v\tan vdv=dx$
$\Rightarrow -\sec v=x+c$
$\Rightarrow x+\sec (y-x)=c$